How do you differentiate #-3=xye^(x-y)#?

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Answer 1 · 2015-12-02T23:18:24

#dy/dx=-(y(x+1))/(x(1-y))#

To find the implicit derivative, we'll use the chain rule and product rule. It's important to remember that when differentiating a #y# term, a #dy/dx# term will be spit out thanks to the chain rule.

#d/dx[-3=xye^(x-y)]#

#0=ye^(x-y)d/dx[x]+xe^(x-y)d/dx[y]+xyd/dx[e^(x-y)]#

Find each derivative.

#d/dx[x]=1#

#d/dx[y]=dy/dx#

#d/dx[e^(x-y)]=e^(x-y)d/dx[x-y]=e^(x-y)(1-dy/dx)=e^(x-y)-e^(x-y)dy/dx#

Plug them back in and solve for #dy/dx#.

#0=ye^(x-y)+xe^(x-y)dy/dx+xye^(x-y)-xye^(x-y)dy/dx#

#-xye^(x-y)-ye^(x-y)=dy/dx(xe^(x-y)-xye^(x-y))#

#(-xye^(x-y)-ye^(x-y))/(xe^(x-y)-xye^(x-y))=dy/dx#

#dy/dx=(e^(x-y)(-xy-y))/(e^(x-y)(x-xy))#

#dy/dx=-(y(x+1))/(x(1-y))#